a space is compact iff any family of closed sets having fip has non-empty intersection

Theorem. A topological space is compact if and only if any collection of its closed sets having the finite intersection property has non-empty intersection.

The above theorem is essentially the definition of a compact space rewritten using de Morgan’s laws. The usual definition of a compact space is based on open sets and unions. The above characterization, on the other hand, is written using closed sets and intersections.

Proof. Suppose $X$ is compact, i.e., any collection of open subsets that cover $X$ has a finite collection that also cover $X$. Further, suppose ${\{F_i\}}_{i\in I}$ is an arbitrary collection of closed subsets with the finite intersection property. We claim that ${\cap }_{i\in I}{F}_i$ is non-empty. Suppose otherwise, i.e., suppose ${\cap }_{i\in I}{F}_i=\mathrm{\varnothing }$. Then,

$X$ $=$ ${\left({\displaystyle \bigcap _{i\in I}}{F}_i\right)}^c$

$=$ ${\displaystyle \bigcup _{i\in I}}{F}_i^c.$

(Here, the complement of a set $A$ in $X$ is written as $A^c$.) Since each $F_i$ is closed, the collection ${\{F_i^c\}}_{i\in I}$ is an open cover for $X$. By compactness, there is a finite subset $J\subset I$ such that $X={\cup }_{i\in J}{F}_i^c$. But then $X={({\cap }_{i\in J}{F}_i)}^c$, so ${\cap }_{i\in J}{F}_i=\mathrm{\varnothing }$, which contradicts the finite intersection property of ${\{F_i\}}_{i\in I}$.

The proof in the other direction is analogous. Suppose $X$ has the finite intersection property. To prove that $X$ is compact, let ${\{F_i\}}_{i\in I}$ be a collection of open sets in $X$ that cover $X$. We claim that this collection contains a finite subcollection of sets that also cover $X$. The proof is by contradiction. Suppose that $X\ne {\cup }_{i\in J}{F}_i$ holds for all finite $J\subset I$. Let us first show that the collection of closed subsets ${\{F_i^c\}}_{i\in I}$ has the finite intersection property. If $J$ is a finite subset of $I$, then

${\displaystyle \bigcap _{i\in J}}{F}_i^c$ $=$ ${\left({\displaystyle \bigcup _{i\in J}}{F}_i\right)}^c\ne \mathrm{\varnothing },$

where the last assertion follows since $J$ was finite. Then, since $X$ has the finite intersection property,

$\mathrm{\varnothing }$ $\ne$ ${\displaystyle \bigcap _{i\in I}}{F}_i^c={\left({\displaystyle \bigcup _{i\in I}}{F}_i\right)}^c.$

This contradicts the assumption that ${\{F_i\}}_{i\in I}$ is a cover for $X$. $\mathrm{\square }$

Title	a space is compact iff any family of closed sets having fip has non-empty intersection
Canonical name	ASpaceIsCompactIffAnyFamilyOfClosedSetsHavingFipHasNonemptyIntersection
Date of creation	2013-03-22 13:34:10
Last modified on	2013-03-22 13:34:10
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	20
Author	CWoo (3771)
Entry type	Theorem
Classification	msc 54D30